internal surface KFGH will not balance each other. To find the displacing force acting on this surface in the direction HF, we may draw a projection of the surface on a plane at right angles to HF, as in Fig. 16a; and subtracting as before the negative from the positive projection, we have the two ellipses, whose minor axes KH and HG will each be equal to Exactly the same result will, of course, be reached if we regard the cylindrical columns AB and BC as members of a fluid polygonal frame under an axial compressive force in each member, equal to P or pd2. 4 When the straight lengths of pipe are formed with plain ends united by double sockets, whose faces are inclined to each other at the angle 0, as sketched in Fig. 17, there will evidently be no unbalanced force resulting from the internal pressure on the sides of the straight pipes; while the socket itself will be subjected to a displacing force in the direction HF, whose magnitude will be independent of the length of the socket, but will be proportional to sin as before. 0 2 It seems probable, however, that the fluid would find its way into the joint so far as to take effect over the annular surface at the ends of the plain pipe; and if their external diameter is d1, the area of each elliptical projection A π will be d2 sin 2; or, if we denote by P1 the axial pressure taken over the whole section of the pipe (including the annular surface), the displacing force will be FIG. 17 F (4α) EXAMPLE. Suppose a cast-iron pipe, having an internal diameter of 36 inches, and an external diameter of 39 inches, to be laid in straight lengths, whose centre lines form the tangents to a curve of 50 feet radius. And let the length of each side, AB, BC, etc., be 5.24 feet, so that the angle subtended at the centre of the curve by each side of the polygon is 6°, the internal angles ABC, etc., being each equal to 174°. If this pipe were charged under a hydrostatic head h of 320 feet, the axial fluid stress would be-- Р = 320 × 62.4 × 32 × 0·7854 ÷ 2240 = 63 tons But, in the case of the socket-joint, the axial stress over the whole area of the pipe would be P1 = 320 × 62·4 × 3·252 × 0·7854 ÷ 2240 = 73.94 tons 1 Also we have tan = 0.0524; sin = 0.05233 The displacing force at right angles to each side of the polygon, in such a case as Fig. 15, would then be which is equivalent to S = 2 × 63 × 0.0524 = 6.7 tons 1.26 tons per lineal foot of the pipe AB. Or, again, at each angle of the same polygon, the radial displacing force would be N: = 2 x 63 x 0.05233 = 6.59 tons But, in the case of the double socket illustrated in Fig. 17, if we allow for the pressure upon the annular ends, we shall have— The displacing force would be equilibrated either by a force equal and opposite to S acting upon each side, or by a force equal and opposite to N acting upon each angle. In this example the difference between N and S is very slight, because is a small angle; but if 0 were large the difference would be greater. Thus, if were 90°, we should have S = N√2. Art. 10. Displacing Forces in a Circular or Segmental Bend. If a cast-iron pipe is formed as a curved bend, its centre line ADB lying in the arc of a circle as sketched in Fig. 18, it is again evident that the pressures on the opposite internal sides of the pipe will not balance each other; for if we conceive the pipe to be divided. into two bent troughs by a diametral section passing along the centre line ADB, the areas of those trough surfaces will not be equal, nor will their projections be equal; and the sum of all the components of fluid pressure measured in the direction OD will be greater than the components measured in the opposite direction. Considering the segment ADB as forming either the entire bend, or one of a series of similar segments constituting the bend, let denote the angle subtended by the arc, and let the segment be terminated at A and B by the radial plane sections OA and OB. Then, to find the magnitude of the displacing force which acts in the direction OD at right angles to the chord AB, draw a projection of the trough surfaces on a plane parallel to AB, and, subtracting the negative area as before, we have the two ellipses of Fig. 18a, in each of which the minor axis is A 2 The combined area of these two ellipses will be 2d2. sin 4 The displacing force in the direction OD, or the resultant of the unbalanced pressures on the internal surfaces of the bend AB, will therefore have the magnitude If we proceed to subdivide the segmental arc ADB into a number of indefinitely small and equal arcs, it is evident that by repeating the same calculation we shall find upon every one of those little arcs a displacing force acting upon the pipe in a direction radially outwards (from the centre of curvature 0), or at right angles to the curved centre line of the pipe; and on each of these equal elements of the arc, the displacing force will have the same magnitude, so that we have really to consider the force as a uniformly distributed pressure acting normally to the arc at all points; and the intensity of this pressure per lineal foot of the pipe is easily to be found. If is indefinitely small, the chord is sensibly equal to the length of the arc, and from the formula (56) above given, it is evident that the distributed radial force or pressure must have for its intensity per lineal unit of arc the value so that the pipe-bend would be perfectly equilibrated at all points if, along with the internal fluid pressure, it were subjected to an external pressure uniformly distributed along the arc, and acting normally to the curve in a direction towards the Р centre 0, with the intensity per unit of arc-length. 4 Exactly the same result will be reached if we regard the curved column of water as a fluid arch of segmental form in which the compressive stress is known to have the value P=pd; for if the compressive stress in such an arch is to follow the curved "line of pressure," of radius R, we know very well, by the ordinary principles of the equilibrium of arches, that it must be combined with a uniform external pressure (normal to the arch), whose intensity per lineal unit of arcР length must be n = as before. R' EXAMPLE. In a cast-iron pipe, whose internal diameter is everywhere 36 inches, let the segmental bend ADB, subtending an angle of 45°, be formed with a radius A0 of 20 feet; and let the pipe be subjected to a hydrostatic head of 320 feet. Here the axial fluid stress P will have the value already found in a previous example, or— Therefore the bend will be subjected to a uniformly distributed radial displacing force or pressure, whose intensity per unit of arc-length will be Upon the whole bend, the resultant of this distributed force will have the direction OD, and the magnitude— N = 2P sin = 2 x 63 x 0.38268 = 48.2 tons 2 Art. 11. Axial Stress in the Fluid and in the Pipe.-In the preceding articles we have found the displacing forces due to the unbalanced fluid pressures by a direct measurement of those pressures upon the actual internal surfaces of the pipe; but in each case it has been shown that the same result might be more simply obtained by considering the axial stress in the cylindrical column of fluid which fills the pipe. For in every equilibrated arch or polygonal frame there must be a certain adjustment between the so-called "thrust," or compressive axial stress, and the external forces applied to the arch or to the joints of the frame. Of course the same adjustment must always subsist whether this axial stress takes effect wholly in the fluid contents of the pipe, or partly or wholly in the metal of the pipe itself; and therefore if we know in any given case that the pipe is itself subjected to a longitudinal compressive force Q, the effective axial stress P+Q must be substituted for P in calculating the external loads or lateral forces that are required to maintain the equilibrium of the arch or frame. The formulæ already found for the polygonal bend will then become while for the segmental bend or arch we shall have (7) (8) (9) or for the uniformly distributed pressure per lineal unit of arc-length n = P+Q (10) EXAMPLE. Suppose the segmental bend ADB already illustrated in Fig. 18 to be fitted at each end with an expansion joint, in which the plain cylindrical end of the pipe slides in the gland of a stuffing-box, as sketched in Fig. 19. Then, if the internal diameter d is uniform throughout the pipe, it is evident that the gross or effective axial stress will be somewhat greater than pd2 on account 4 π of the fluid pressure upon the annular end of the pipe within the expansion joint. π If d, denotes the external diameter of the sliding neck, or the bore of the gland, the area of annular surface will be (d ̧2 — d2)4, and the total axial stress P+ Q will be equal to pd2. ト。 For example, let d = 36 inches, di = 39 inches, R = 20 feet, and 0 = 45°, as in the previous example; then for a head of 320 feet, we have as before P = 63 tons, P + Q 73.94 tons; and the total lifting-force exerted in the direction OD will be = N = 2 x 73.94 x 0.38268 = 56.59 tons |